elementary statics and dynamics
TRANSCRIPT
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CBSE CLASS XII MATHSElementary Statics And Dynamics
Two mark questions with answers
Q1.To a man walking at the rate of 8 km h-1, rain appears to fall vertically
downwards. find the actual direction of the rain if its actual velocity is 82 km h-1.
Ans1.Let the man e moving along !" at the rate of 8 km h-1and let the rain
appear to fall along !#$. Let the rain make angle with !#. %ain is falling with avelocity of 82 kmh-1.
82 kmh-1is the resultant of 8 kmh-1and &%'.%esolving along !"82.cos(2)* + 8cos+ &%'.cos2)*o
82sin 8 &%'velocity rain /.r.t man0sin 12 3o.
Q2.Two forces of magnitudes 24 and 4 act an a point and the angle etween them
is 5*o. 6ind the magnitude and the direction of their resultant.
Ans2.%esultant for the forces 7 and inclined at an angle is
%esultant (22+ 2+ 2x2x cos5*9 ( + 15 + 8 2) 4:f the resultant makes angle with 7;irection of resultant is tan sin(7 + costan (2xsin5*o(2 + xcos 5*o0
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these stringsF
Ans".Let T and T$ e the two tensions. =ince the mass is in equilirium, the
resultant of the tensions must e equal and opposite to the 1* kg w.t.
%esolve the 1*kg w.t. along the lines of action of the tension i.e., along lines makingangles
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/e have, T and % .
Therefore,
tan
.
Q11.=how that the path of a particle moving in space with constant acceleration is aparaolaF
Ans11.Ccceleration is constant.
constant a (say.
:ntegrating, we get aI + .
Cgain integrating, we get y , which is a paraola.
Q12.C truck is moving along a level road at the rate of )2 km per hour. :n whatdirection must a ullet e shot from it with a velocity of
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&G.
&G K
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Ans1.=uppose the forces 7 and act at an angle and let their resultant e %.Then
%2 72+ 2+ 27 cos .....(i:t is given that % is also the resultant of the forces 7 + = and - = acting at the
same angle. Therefore,
%2 (7 + =2+ ( - =2+ 2(7 + =( - = cos ....(ii6rom (i and (ii, we get
72+ 2+ 27 cos (7 + =2+ ( - =2+ 2(7 + = ( - = cos * 2=2+ 2=(7 - + 2=( - 7 cos - 2=2cos * 2=K(= + 7 - - (= + 7 - cos (= + 7 - cos (= + 7 -
cos 1 ...= - 7 ... (= + 7 - *07utting cos 1 in (i, we otain,%2 72+ 2+ 27 (7 + 2
% 7 + >ence, the magnitude of the resultant of 7 and is 7 + .
Q2.Two forces each of magnitude 7 act at a point. 6ind the angle etween tham ifthe magnitude of their resultant is equal to (1 7 and (2 72.
Ans2.Let e the angle etween the two equal forces each of magnitude 7. Then,their resultant % is given y
% 27cos(2 .......(sustituting the values in % (72+ q2+ 2pq cos(1 :f the resultant of the equal forces is 7, then,
% 27 cos(27 27 cos(2 ...% 701 2 cos(2cos(2 12 cos(2 cos5*92 5*9 12*9>ence, the angle etween the forces is 12*9.
(2 :f the resultant of the equal forces is 72, then,
% 27 cos(272 27 cos(2cos(2 12 cos-1(1 2cos-1(1 cos-1K2x(12- 1 ...2cos-1I cos-1(2I2- 10 cos-1(-)8>ence, the angle etween the forces is cos-1(-)8.
Q3./ith two forces acting at a point, the maIimum effect is otained when their
resultant is 4ewtons. :f they act at right angles, their resultant is < 4ewtons. 6indthe forces.
Ans3.Let the two forces e of magnitudes 7 and 4ewtons. /e know that resultant
(% of two forces (7 and is maIimum when cosis maIimum. Gut maIimum valueof cos 1, when * then,%2 72+ 2+ 27.1
%2 (7 + 2
% 7 +
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cos N(
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< sin (7 + cos< sin 7 + (22- 72270 ....using (i0< sin 7 + (22- 72270< sin (72+ 222sin (72+ 2227 < ......(ii4ow, cos2+ sin2 1(22- 722702+ (72+ 2227
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&72- u2 2gh -------(2
u2- 2gh0 - u2 2gh
u2- 8gh - u2 2gh
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Ans3.Let the particle e proJected from C and let it returns to the ground aG. Let 7
e the position of the particle at some instant moving with velocity u at angle withthe horiDontal. Let after time t. The particles reaches where its velocity is v O
makes an angle (?*P - to the horiDontal component of velocity constantthroughout the motion.
>oriDontal component of velocity of 7 horiDontal component of velocity at ucos vcos (?*P - ucos vsin....................(iClso, for vertical of velocity, using equation
v u - gt, for upward motioninitial velocity is usinand final velocity eing - vcos.-vcos usin- gtgt usin+ vcos.................(ii7utting the value of v from (i in (iigt usin+ ucossinxcosgt u (sin2+ cos2sin usingt u cosect ug cosec>ence, the particle will e at right angle to the previous direction after time ug
cosec.
Q4.:f v1O v2e the velocities at the ends of a focal chord of a proJectile path O u e
the velocity at the verteI of the path. Then show that 1v12+ 1v2
2 1u2.
Ans4.
Let in figure 7 e the given focal chord of the paraolic traJectory followed y aparticle proJected from !. /e know that the tangents drawn at the end points of a
focal chord intersect at right angles. ... :f v1is the velocity of the particle at 7 makingan angle with the horiDontal 7I$. Then velocity v2of the particle at will make anangle (?*P - to the horiDontal. Let C e the verteI of the paraola where velocity
of the particle is u, directed horiDontally.... >oriDontal component of velocity remains constant throughout the motion.... >oriDontal component of velocity at 7 velocity at C horiDontal component of
velocity at v1cos u v2cos(?*P - v1cos u v2sintaking first two terms, we get
cos uv1taking second two terms, we get
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sin uv2=quaring and adding (i O (ii
cos2+ sin2 u2v12+ u2v22
1u2 1v12+ 1v2
2.
Q.C particle is proJected from a point at an angle and after time t is oserved to
have an elevation as seen from the point of proJection prove that the initial velocitywas (gt cos(2sin(- .
Ans.Let the particle e proJected from point ! with velocity u at an angle withthe horiDon !I. Let the particle e at 7 after time t. =uch that !7 E O I!7 .
;raw perpendicular 7C on !I from 7, then in !C7, 7C E sin O !C Ecos
:n time t, horiDontal O vertical distances transverse y the particle are !C O 7C.4ow the horiDontal O vertical components of the initial velocity are ucosO usin.The horiDontal distance covered y the particle in time t is !C
!C (ucostEcos ucos.t .....................(iThe vertical distance coverd y the particle in time t is 7Ch ut - (12 gt2
7C (usint - gt2
Esin usin.t - (12gt2....................(ii;ividing (i y (ii, we get(Ecos(Esin (ucos.tusin.t - (12get20cos2usin- gt0 2ucossin
2usincos- cossin0 gt cosu (gtcos2sin(- >ence the initial velocity (gtcos2sin(- 0.
Q!.C particle is proJected with a velocity 2ag. =o that it Just clears two walls ofequal heights a which are at a distance 2a from each other. =how that the latus
rectum of the traJectory is equal to 2a and the time of passing etween the walls is a
2ag
Ans!.Let C' and G4 e two walls of equal heights a at a distance 2a. CG '4 2a
4ow, we know that the velocity of a proJectile at any point in its path is equal to the
velocity of free fall from the directriI to that point.>ight of the ditrectriI aove thepoint of proJection
u22g u 2gh h u22g0
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(u sin t - (12gt2u0 cos . t ....... KMsing (ii and (iii (2usin - gt2ucos 2usin - (u sin - v sin02u cos ....... KMsing (iv u sin + v sin2u cos u sin + (u coscos . sin02u cos ...... KMsing (i 12 tan + tan0>ence, 7roved